With a 500hp shunt wound DC motor, at 1800rpm, 500Vdc, FLA is 800Adc, Ra is 0.02

Question

With a 500hp shunt wound DC motor, at 1800rpm, 500Vdc, FLA is 800Adc, Ra is 0.025 and Rf is 400.   I have the following, Armature: current, resistance and volts, Field: current, resistance and volts, Volts per rpm, Power in and out, and torque.  I am sort struggling with a few parts,

First would Armature current be 800A? I hope so as that is what I used to find some of the above info.  

What would Va need to be if you wanted the motor to operate at 1000rpm at rated torque, keep field current constant.

So for that would I use Va =  Eg + Ia(Ra), where Eg = (500V / 1800rpm) * 1000rpm?  The other question is how do I find the no load speed?  

Thanks for the help

Explanation / Answer

for dc shunt motor: Va = Eg + Ia(Ra)

given : for 1800 rpm, Ia=800 A, Ra= 0.025 ohms

Va= 500 volts for 1800 rpm

Eg=Va-IaRa= 500-(800*0.025)=500-20=480 volts

also Eg=( flux*Z*N*P)/60*A

Implies:: Eg1/Eg2=N1/N2

480/Eg2= 1800/1000

Eg2=266.67 volts for 1000rpm

here torque is constant:: Ia constant

Va = Eg2+IaRa=266.67+20=287.67 volts for 1000 rpm

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